Numeration Systems and Whole Number Operations
3.2 Addition of Whole Numbers
Models for Addition
Now that we're a little more familiar with how to represent a number in different systems, specifically whole numbers, we're going to look at our first operation, additition.
Recall that whole numbers include 0.
$$\mathbb{W}=\{0,1,2,3,4,\ldots\}$$
At a basic level, addition means combining amounts. You can show that with sets, number lines, blocks, or algorithms. A formal set definition of addition might look something like this.
Set Model for Addition
Definition: Set Model of Addition
If $A$ and $B$ are disjoint finite sets with $n(A)=a$ and $n(B)=b$, then
$$ a+b=n(A\cup B) $$
Note here that the sets are required to be disjoint (why?). The set property we're using is the cardinal number of a set, which is simply counting how many elements are in that set. We've defined addition of two whole numbers to be the cardinal number of the union of the two sets who correspond to those cardinal numbers. In the simplest example, consider the two sets $A$ and $B$.
$$A=\{s\}$$ $$B=\{t\}$$
The cardinal number of $A$ is $1$, it has one thing in it, and the cardinal number of $B$ is $1$. $A$ and $B$ are also disjoint, meaning they have nothing in common, $A\cap B = \{\}$. If we union the sets together,
$$A\cup B=\{s,t\}$$
we get a new set with a cardinal number of $2$. In our formal notation,
$$n(A)=1,n(B)=1,n(A\cup B)=2$$
which directly translates to $$1+1=2$$
Example: Set model
Use the set model to find $7+2$.
Show Solution
To begin, we need two disjoint sets. One of which has a cardinal number of 7, and the other that as a cardinal number of 2. To be a little less formal, this might be presented something like:
Johnny has $7$ apples and Lisa has $2$ apples. When they combine their apples together, how many apples do they have? Counting we can see 9 apples, so $7+2=9$.
This is anagolous to our set model for addition, if $A$ is the set of apples belonging to Johnny and $B$ is the set of apples belonging to Lisa, then we have two disjoint sets. The union of the sets would be the set of apples belonging ot Johnny or Lisa:
$$7+2=n(A\cup B)=9$$
MyOpenMath: Set Model for Addition
Number Line Model for Addition
Another great way to model addition is to use our natural instinct to measure distances. A number line is a line drawn with numbers marked on it, preferrably with evenly spaced intervals for ease of reading. We can visualze a whole number by drawing an arrow on that number line starting at zero and ending at the whole number we want to represent.
Addition on the numberline can be modeled by drawing two numbers, one starting at zero and the second starting at the tip of the first. Where the combined arrow ends is the sum of the two numbers.
Definition: Number Line Model of Addition
To find $a+b$, start at 0, move right $a$ units, then right $b$ more units.
Example: Number line model
Find $7+5$ on a number line.
Show Solution
Start at 0, move to 7, then move 5 more steps:
$$ 7\to8\to9\to10\to11\to12 $$
So $7+5=12$.
Common mistake: starting at 1. That gives the wrong length.
MyOpenMath: Number Line Model for Addition
Ordering and Comparison
With a few models under our belt, its time to look at some properties of whole numbers and in turn properties of addition. First, whole numbers have a relationship in that if you pick any two whole numbers, say $a$ and $b$, then they fall into exactly one of three categories.
$a$ is larger than $b$, $a$ is smaller than $b$, or $a$ is the same as $b$.
This is called a trichotomy, and it is true for all whole numbers. We shorthand these relationships with symbols.
Definition: Less Than and Greater Than (Whole Numbers)
For any whole numbers $a,b$, exactly one of the following is true:
- $a<b$ means there exists $k\in\mathbb{N}$ with $a+k=b$.
- $a>b$ means there exists $k\in\mathbb{N}$ with $b+k=a$.
- $a=b$
Note that I'm defining the notation here in reference to something we already understand, addition. This is neccessary if we want to build up new operations and algorithms and still be sure that they hold. Everything here is defined on addition, for which we already have 2 models to justify.
We can confidently say $2<3$, because based on our definition we can find a natural number $k$ such that $2+k=3$ That $k$ happens to be $1$. because $2+1=3$, $2$ is definitely less than $3$.
Our definition will not hold if we allow $k$ to be any whole number, because that means it is possible that $k=0$. That would allow for something like $2+0=2$ implying that $2$ is less than $2$, but that is clearly not the case. $2=2$. This relationship falls in that third category of the trichotomy.
Example: Justify the inequality
Using our definition above, decide if the statements below are true. If so, justify the statement.
- $3<5$
- $9>4$
- $16<17$
- $3<3$
Show Solution
- $3+2=5$, so $3<5$.
- $4+5=9$, so $9>4$.
- $16+1=17$, so $16<17$.
- $3<3$ is false because the only number that solves $3+k=3$ is $k=0$ which is not a natural number.
MyOpenMath: Justify inequality
Properties of Whole Number Addition
Here's a whole list of things that are true about whole numbers and addition.
Fact: Properties of Whole Number Addition
For any whole numbers $a,b,c$ the following hold:
- Closure: $$a+b\in\mathbb{W}$$
- Commutative: $$a+b=b+a$$
- Associative: $$(a+b)+c=a+(b+c)$$
- Identitiy Element: $$a+0=a$$
These properties may seem daunting at first, but they are very basic facts about whole number addition. These are the rules that give us the confidence to make larger statements about numbers in general. The individual numbers themselves are not important here, their behavior is important.
Closure means that no matter the whole numbers you pick, when you add them together you will always get another whole number.
$$3+4=7$$ $$500+123=623$$ $$10^{400}+1 = 10000\ldots 001$$
A mathematician would say that the whole numbers are closed under addition. This is an important distinction because not all sets are. Take the set of odd whole numbers for example, $$S=\{1,3,5,7,9,\ldots\}$$
Picking any two elements of that set would not stay in the set! $$1+3=4\notin S$$
MyOpenMath: Decide if set is closed under addition
The commutative property says it does not matter the order in which you add the numbers. $2+3$ will always be the same as $3+2$. We can demonstrate this with either of our addition models above. Heres the number line again, use it to show that $2+3$ yields the same value as $3+2$
The associative property tells us that if we're adding multiple numbers together, not just 2, it doesn't matter the way you group the numbers.
$$2+(3+4)=(2+3)+4$$
Using our set model, let $A,B,C$ be three disjoint sets.
$$A=\{a,b\}$$ $$B=\{c,d,e\}$$ $$C=\{f,g,h,i\}$$
If we first combine $A\cup B$, we get
$$A\cup B = \{a,b,c,d,e\}$$
Then adding on $C$
$$(A\cup B)\cup C=\{a,b,c,d,e,f,g,h,i\}$$
which tells us that $(2+3)+4=9$. If instead we grouped $B\cup C$ first,
$$B\cup C = \{c,d,e,f,g,h,i\}$$
and then added in $A$,
$$A\cup (B\cup C) = \{a,b,c,d,e,f,g,h,i\}$$
yields exactly the same set even though we calculated $2+(3+4)$. Neither the order nor the grouping matters for addition.
The additive identity is an element in the set that when added to another element, the value doesn't change. The fact that whole numbers have $0$ is what sets them apart from the natural numbers. We saw how important having $0$ is when we looked at numeration systems that didn't have one like the Babylonian's cuneiform.
Example: Name the property
- $9+8=8+9$
- $57+999\in\mathbb{W}$
- $3+(4+5)=(3+4)+5$
- $(8+5)+2=2+(8+5)$
Show Solution
- Commutative
- Closure
- Associative
- Commutative. The grouping didn't change, what changed here was the order of the terms.
MyOpenMath: Naming Property of Whole Number Addition
Addition Algorithms
Now its time to start looking at ways to add numbers together efficiently. We're not going to count out whole sets of numbers every time we need to do an addition! There are algorithms that we can use which involve manipulating numerals.
Standard Algorithm
The first of which we will talk about is the standard algorithm. But before we dive into it directly, lets look at a learning tool that shows up in classrooms which the algorithm mimics. Students are given blocks that snap together to visualize adding numbers. Heres a short video from a first grade classroom.
Things start to get weird around the edge case of 10 blocks. Because we only have 10 symbols to represent number (in base 10), once we get to 10 blocks we set them to the side and refer to that grouping as a 10 and continue to count. Heres a good video showing that process.
Use this Desmos plot to try out the blocks in a digital space.
The standard algorithm mimics this by first lining digits up and then adding straight down. Lets use an example to get a better idea.
Example: Standard algorithm of addition
Compute $186+127$.
Show Solution
The algorithm begins by lining up the numbers by their place vertically and adding straight down starting with the right most column.
| op | hundreds | tens | ones |
|---|---|---|---|
| 1 | 8 | 6 | |
| + | 1 | 2 | 7 |
We add the ones place. $6+7=13$, so we write 3 and carry the 1. This is representative of snapping 10 of the 13 units together and moving it to the pile of longs.
| op | hundreds | tens | ones |
|---|---|---|---|
| 1 | 1+8 | 6 | |
| + | 1 | 2 | 7 |
| 3 |
Now, adding the tens, or the longs place, we have $1+8+2=11$. So we carry the 1 and write the 1 in the longs place. This is snapping 10 longs into a flat and moving it to the pile of flats.
| op | hundreds | tens | ones |
|---|---|---|---|
| 1+1 | 1+8 | 6 | |
| + | 1 | 2 | 7 |
| 1 | 3 |
Finally, we add the hundreds, or the flats place, and we have $1+1+1=3$. There is no need to carry here so we can write the $3$ down and read off our answer.
| op | hundreds | tens | ones |
|---|---|---|---|
| 1+1 | 1+8 | 6 | |
| + | 1 | 2 | 7 |
| 3 | 1 | 3 |
Therefore $$186+127=313$$
Base-5 Addition
Now its time to flex our skills by working in a different base. If the rules to our little block game were not to snap 10 units together to get a long, and instead they were snap only 5 together to make a long, then we would be counting in base 5. As strange as this upcoming problem might be, know that it is the same algorithm. Only that our rules have changed. We no longer have 10 symbols to work with
$$D_{10}=\{0,1,2,3,4,5,6,7,8,9\}$$
but only
$$D_{5}=\{0,1,2,3,4\}$$
When starting out with addition, it can be helpful to have a table of all of the possible sums. For example, to know what $7+8$ is, we coudl go to the 7th row and the 8th column of an addition table in base 10 to see the number is $15$.
| + | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 2 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 3 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| 4 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
| 5 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
| 6 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| 7 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
| 8 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| 9 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
A base-5 addition table looks like this. To find a sume of say $3+4$ we would go to the 3rd row and the 4th columna nd see the value is $12_{\text{five}}$. Remember that in base 5, that is $$1(5)+2(1)=7_{\text{ten}}$$
| + | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 3 | 4 |
| 1 | 1 | 2 | 3 | 4 | 10 |
| 2 | 2 | 3 | 4 | 10 | 11 |
| 3 | 3 | 4 | 10 | 11 | 12 |
| 4 | 4 | 10 | 11 | 12 | 13 |
MyOpenMath: Build your own addition table
Example: Base-5 Addition
Compute $4_{\text{five}}+1_{\text{five}}$ in base five
Show Solution
In base 10, $4+1=5$. But in base 5, 5 is written as one five and zero ones: $$1(5)+0(1)=5$$
$$4_{\text{five}}+1_{\text{five}}=10_{\text{five}}$$
Check the addition table above to verify.
Example: Base-5 Addition
Compute $3144_{\text{five}}+434_{\text{five}}$
Show Solution
Using the standard algorithm, line them up.
| op | 125's | 25's | 5's | 1's |
|---|---|---|---|---|
| 3 | 1 | 4 | 4 | |
| + | 4 | 3 | 4 | |
Adding in the 1's column we have $$4+4=8=13_{\text{five}}$$
That means we write down a 3 and carry the 1 (but what we're actually carrying is a group of 5).
| op | 125's | 25's | 5's | 1's |
|---|---|---|---|---|
| 3 | 1 | 1+4 | 4 | |
| + | 4 | 3 | 4 | |
| 3 |
Next up, the 5's column. $$1+4+3=8=13_{\text{five}}.$$ Again we write down the 3 and carry the 1 (this time we're carrying a group of 25's).
| op | 125's | 25's | 5's | 1's |
|---|---|---|---|---|
| 3 | 1+1 | 1+4 | 4 | |
| + | 4 | 3 | 4 | |
| 3 | 3 |
Now the 25's column, $$1+1+4=6=11_{\text{five}}$$ Mark down a 1 and carry the 1 (125).
| op | 125's | 25's | 5's | 1's |
|---|---|---|---|---|
| 1+3 | 1+1 | 1+4 | 4 | |
| + | 4 | 3 | 4 | |
| 1 | 3 | 3 |
$1+3=4=4_{\text{five}}$. Write down the 4 and we're done!
| op | 125's | 25's | 5's | 1's |
|---|---|---|---|---|
| 1+3 | 1+1 | 1+4 | 4 | |
| + | 4 | 3 | 4 | |
| 4 | 1 | 3 | 3 |
This shows
$$3144_{\text{five}}+434_{\text{five}}=4133_{\text{five}}$$
MyOpenMath: Addition in a base other than 10
The Lattice Algorithm
The lattice algorithm is another algorithm for addition. It is similar to the standard algorithm in that we line the numbers up, but instead of carrying to the next column after a sum, we write the number vertically. When we have finished each sum, we then sum the diagonals of the lattice that has been constructed.
Example: Lattice addition
We're in base 10. Compute $1507+383$.
Show Solution
Write $1507$ and $383$ stacked, lining up the places. And below we'll write the sums, but instead of carrying we'll stack the numbers.
| op | 1000's | 100's | 10's | 1's |
|---|---|---|---|---|
| 1 | 5 | 0 | 7 | |
| + | 3 | 8 | 3 | |
We start with $7+3=10$, but instead of carrying that 1, we'll just write the number vertically.
| op | 1000's | 100's | 10's | 1's |
|---|---|---|---|---|
| 1 | 5 | 0 | 7 | |
| + | 3 | 8 | 3 | |
| 1 | ||||
| 0 |
We do this for each column.
| op | 1000's | 100's | 10's | 1's |
|---|---|---|---|---|
| 1 | 5 | 0 | 7 | |
| + | 3 | 8 | 3 | |
| 0 | 0 | 0 | 1 | |
| 1 | 8 | 8 | 0 |
And now we add along the diagonals, starting with the smallest value at the very bottom right.
| op | 1000's | 100's | 10's | 1's |
|---|---|---|---|---|
| 1 | 5 | 0 | 7 | |
| + | 3 | 8 | 3 | |
| 0 | 0 | 0 | 1 | |
| 1 | 8 | 8 | 0 | |
| 0 |
For the next diagonal $8+1=9$
| op | 1000's | 100's | 10's | 1's |
|---|---|---|---|---|
| 1 | 5 | 0 | 7 | |
| + | 3 | 8 | 3 | |
| 0 | 0 | 0 | 1 | |
| 1 | 8 | 8 | 0 | |
| 9 | 0 |
And the third, $0+8=8$.
| op | 1000's | 100's | 10's | 1's |
|---|---|---|---|---|
| 1 | 5 | 0 | 7 | |
| + | 3 | 8 | 3** | |
| 0 | 0 | 0 | 1 | |
| 1 | 8 | 8 | 0 | |
| 8 | 9 | 0 |
And for the final one.
| op | 1000's | 100's | 10's | 1's |
|---|---|---|---|---|
| 1 | 5 | 0 | 7 | |
| + | 3 | 8 | 3 | |
| 0 | 0 | 0 | 1 | |
| 1 | 8 | 8 | 0 | |
| 1 | 8 | 9 | 0 |
It must be the case that $$1507+383=1890$$
MyOpenMath: Lattice Addition
Mental Strategies for Addition
Fact: Mental Strategies
Useful mental strategies: counting on, doubles, making 10, compatible numbers, and front-end estimation.
Partial Sums
Because of our commutatitve and associative property for addition, it really doesn't matter the order or grouping you use when adding whole numbers. We can use this to our advantage by breaking complicated sums into easier expressions.
Example: Partial Sum
Compute $67+36$.
Show Solution
At first glance you may want to run to an algorithm we've talked about here to add these numbers, but theres an easier way.
$$67=60+7$$ $$36=30+6$$
Instead, we can do the sum $$60+7+30+6$$
Grouping the tens together seems like a nice move. $60+30=90$
$$60+30+7+6$$
$$90+7+6$$
Now we can add the ones and finish the sum.
$$90+13=103$$
Using Doubles
People tend to be very good at doubling numbers. You may have looked at that last partial sum and saw the $7+6$ and think to your self, thats just 1 more than $6+6$. If so, then you're already using doubles. Splitting sums into pieces that include doubles is a great way to help with mental arithmetic.
Example: Using Doubles
Evaluate $31+44$
Show Solution
At first glance this doesn't seem particlarly double-y. But If we take the 1 off of the 3 and 14 off of the 44 we have
$$30+1+30+14$$ $$30+30+15$$ $$60+15$$ $$75$$
Not a bad way to think about the problem.
Breaking and Bridging
In truth, we've already been breaking sums into smaller pieces, but we need not aim for doubles.
Example: Breaking and bridiging
Evaluate $67+36$
Show Solution
If we borrowed 3 from the 36 to give to 67, it would become a nice round 70.
$$67+33+3$$ $$70+33$$
And then we could break off 3 more from the 33 to get a nice round 100.
$$70+30+3$$ $$100+3=103$$
Using Compatible Numbers
When doing large sums, it can be very helpful to regroup the terms with terms that are compatible, which is just a nice way of saying easy to calculate.
Example: Compatible Numbers
Evaluate $$130+50+70+20+50$$
Show Solution
Right off the bat I see two 50's which is a nice 100. And the $130+70=200$. So shuffling the sum around we get
$$130+70+50+50+20$$ $$200+100+20$$ $$320$$
Counting On
If the numbers are small enough, you will often hear cashiers counting back cash to find sums. For example, if I payed for a \$13 item with \$20 bill, the cashier might hand back two \$1 bills, counting fourteen, fifteen, and then a \$5 bill, counting twenty.
Thats because the cost of the item, plus what the cashiere gives back to me, should add up to the \$20 that was given.
$$13+1+1+5=20$$
If I had paid with a \$50, the count back may have gone like this
$$13+1+1+5+20+10=50$$
Counting up to these round and easy to use numbers is a great way to do addition.